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2 years ago

What are the 3 ways that vectors can be found in books?

1 answer:

6 0

I'm not quiet sure what exactly this question is asking but some examples of vectors include displacement, velocity, position, force, and torque.

You might be interested in

A car travels 200km in 3.0 hours. Determine the average velocity of the car

Masja [62]

Answer:

Explanation:

The average velocity of the car can be found by using the formula

$a = \frac{d}{t } \\$

d is the distance

t is the time taken

From the question we have

$a = \frac{200}{3} \\ = 66.66666...$

We have the final answer as

Hope this helps you

4 0

3 years ago

Astronomers know that the distance between the Earth and the Sun averages 1.50 x108 km. How can astronomers use the observed ste

rodikova [14]

Answer:

The distance of stars and the earth can be averagely measured by using the knowledge of geometry to estimate the stellar parallax angle(p).

From the equation below, the stars distances can be calculated.

D = 1/p

Distance = 1/(parallax angle)

Stellar parallax can be used to determine the distance of stars from an observer, on the surface of the earth due to the motion of the observer. It is the relative or apparent angular displacement of the star, due to the displacement of the observer.

Explanation:

Parallax is the observed apparent change in the position of an object resulting from a change in the position of the observer. Specifically, in the case of astronomy it refers to the apparent displacement of a nearby star as seen from an observer on Earth.

The parallax of an object can be used to approximate the distance to an object using the formula:

D = 1/p

Where p is the parallax angle observed using geometry and D is the actual distance measured in parsecs. A parsec is defined as the distance at which an object has a parallax of 1 arcsecond. This distance is approximately 3.26 light years

3 0

3 years ago

An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the

OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

$m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2$

Through the aforementioned formula we will have to

$v_f^2-v_i^2 = 2ax$

The particulate part of the rest, so the final speed would be

$v_f^2 = 2gx$

$v_f=\sqrt{2(9.8)(3.1)}$

$v_f = 7.79m/s$

Now from Newton's second law we know that

$F = ma$

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

$F = m\frac{dv}{dt}$

Replacing we have that,

$F = (89)\frac{7.79}{0.5}$

$F = 1386.62N$

Therefore the force that the water exert on the man is 1386.62

3 0

3 years ago

H. A truck starts to move from rest. If it gains the acceleration of 3 m/s2 in 5 sec,

SashulF [63]

Answer:

1. The final velocity of the truck is 15 m/s

2. The distance travelled by the truck is 37.5 m

Explanation:

1. Determination of the final velocity

Initial velocity (u) = 0 m/s

Acceleration (a) = 3 m/s²

Time (t) = 5 s

Final velocity (v) =?

The final velocity of the truck can be obtained as follow:

v = u + at

v = 0 + (3 × 5)

v = 0 + 15

v = 15 m/s

Therefore, the final velocity of the truck is 15 m/s

2. Determination of the distance travelled

Initial velocity (u) = 0 m/s

Acceleration (a) = 3 m/s²

Time (t) = 5 s

Distance (s) =?

The distance travelled by the truck can be obtained as follow:

s = ut + ½at²

s = (0 × 5) + (½ × 3 × 5²)

s = 0 + (½ × 3 × 25)

s = 0 + 37.5

s = 37.5 m

Therefore, the distance travelled by the truck is 37.5 m

6 0

2 years ago

A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torqu

ira [324]

Answer:

$-20.0 kg m^2/s$

Explanation:

The angular momentum of an object in rotation is given by

$L=I \omega$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, initially we have

$I=2 kg m^2$ is the moment of inertia of the wheel

$\omega_i = 6.0 rad/s$ is the initial angular speed

So the initial angular momentum is

$L_i = I\omega_i = (2)(6.0)=12 kg m^2/s$

Later, a counterclockwise torque of

$\tau=-5.0 Nm$ is applied

So the angular acceleration of the wheel is:

$\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2$ in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

$\omega_f = \omega_i + \alpha t$

where

t = 4.0 is the time interval

Solving,

$\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s$

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

$L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s$

So, the change in angular momentum is:

$\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2$

7 0

3 years ago